Problem: $ D = \left[\begin{array}{rrr}5 & 1 & 3 \\ -2 & 0 & 3\end{array}\right]$ $ F = \left[\begin{array}{rr}5 & -1 \\ 4 & 1 \\ 5 & 3\end{array}\right]$ What is $ D F$ ?
Explanation: Because $ D$ has dimensions $(2\times3)$ and $ F$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D F = \left[\begin{array}{rrr}{5} & {1} & {3} \\ {-2} & {0} & {3}\end{array}\right] \left[\begin{array}{rr}{5} & \color{#DF0030}{-1} \\ {4} & \color{#DF0030}{1} \\ {5} & \color{#DF0030}{3}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{5}\cdot{5}+{1}\cdot{4}+{3}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{5}+{1}\cdot{4}+{3}\cdot{5} & ? \\ {-2}\cdot{5}+{0}\cdot{4}+{3}\cdot{5} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{5}+{1}\cdot{4}+{3}\cdot{5} & {5}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{1}+{3}\cdot\color{#DF0030}{3} \\ {-2}\cdot{5}+{0}\cdot{4}+{3}\cdot{5} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{5}\cdot{5}+{1}\cdot{4}+{3}\cdot{5} & {5}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{1}+{3}\cdot\color{#DF0030}{3} \\ {-2}\cdot{5}+{0}\cdot{4}+{3}\cdot{5} & {-2}\cdot\color{#DF0030}{-1}+{0}\cdot\color{#DF0030}{1}+{3}\cdot\color{#DF0030}{3}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}44 & 5 \\ 5 & 11\end{array}\right] $